# Continuity of a function

1. Nov 16, 2009

### utleysthrow

1. The problem statement, all variables and given/known data

f:R2 -> R

f(x,y) = $$e^{-x^{2}/y^{2}}$$ if y is not 0, and 0 if y is 0

a) At (1,1), is f continuous?
b) At (1,0), is f continuous?

2. Relevant equations

The function f is continuous at the point c if for every sequence (xn) in X with limit lim xn = c, we have lim f(xn) = f(c).

3. The attempt at a solution

a) I think it is continuous at (1,1). My reasoning is that $$e^{-x^{2}/y^{2}}$$ is continuous on $$R^{2}-\left\{(x,0)\right\}$$, and $$(1,1) \in R^{2}-\left\{(x,0)\right\}$$.

b) I don't think this is true... but not sure how to show it.

I am trying to work with the definition: "The function f is continuous at the point c if for every sequence (xn) in X with limit lim xn = c, we have lim f(xn) = f(c)."

So if f is continuous at (1,0), that means for any lim (xn,yn) = (1,0), we have lim f(xn,yn) = f(1,0). But f(1,0) is 0, so any lim f(xn,yn) = 0.

If I choose (xn,yn) = (1, 1/n), the limit is (1,0). But lim f(xn,yn) = lim $$e^{-n^{2}}$$, which does not equal 0. Does this show that f is not continuous at point (1,0)?

Last edited: Nov 17, 2009
2. Nov 16, 2009

### Staff: Mentor

For b, if your function is continuous at (1, 0), then lim f(x, y) = f(1, 0) = 0, as (x, y) approaches (1, 0), independent of the path taken to get to (1, 0). If you can find one path into (1, 0) that produces a function value other than 0, that would suffice to show that the limit doesn't exist.

3. Nov 17, 2009

### utleysthrow

Is my example of such a path (that I stated above) not valid?

If (xn, yn) = (1, 1/n), it goes to (1,0), but the limit of the function value, lim f(xn, yn) = lim $$e^{-1/\frac{1}{n^{2}}}$$ = lim $$e^{-n^{2}}$$, which does not equal 0... it goes to infinity as n goes to infinity.

Last edited: Nov 17, 2009
4. Nov 18, 2009

### utleysthrow

okay, I guess it wasn't. f IS continuous at (1, 0), and the function value of any path into (1, 0) is always 0.

For f to be continuous at (1,0), this condition must hold: $$lim_{(x, y)\rightarrow(1, 0)} f(x,y) = f(1,0)=0$$

First, if $$y=0$$, then f(x, y)=f(1, 0)=0. Second, if $$y\neq0$$, since $$(x,y) \rightarrow (1,0)$$, we have $$-x^{2}/y^{2} \rightarrow -\infty$$ and thus $$f(x,y)=e^{-\infty} \rightarrow 0$$. So f must be continuous at (1, 0)

5. Nov 18, 2009

### Staff: Mentor

Yes, I agree. In your previous post, you said that lim f(xn, yn) = lim e-n2, which does not equal zero (limit as n --> infinity for yn = 1/n). As n gets large w/o bound, though, e-n2 approaches 0.